Question

Ajutor. Mulțumesc mult

Answer 1

$a)\\125!=1*2*3*4*...*122*123*124*125=123!*124*125\\123!=1*2*3*4*...*122*123\\\\\frac{125!}{123!}=\frac{123!*124*125}{123!}=124*125=15500\\\\b)\\(n+2)!=1*2*3*...*n*(n+1)*(n+2)\\(n+4)!=1*2*3*...*n*(n+1)*(n+2)*(n+3)*(n+4)=(n+2)!*(n+3)*(n+4)\\\\\frac{(n+2)!}{(n+4)!}=\frac{(n+2)!}{(n+2)!*(n+3)*(n+4)}=\frac{1}{(n+3)*(n+4)}=\frac{1}{n^2+7n+12}\\\\c)\\n!=1*2*3*4*...*(n-1)*n=(n-1)!*n\\(n-1)!=1*2*3*4*...*(n-2)*(n-1)\\\\\frac{n!}{(n-1)!}=\frac{(n-1)!*n}{(n-1)!}=n$\

$d)\\(n+3)!=1*2*3*...*n*(n+1)*(n+2)*(n+3)=(n +1)!*(n+2)*(n+3)\\(n+1)!=1*2*3*...*n*(n+1)\\\\\frac{(n+3)!}{(n+1)!}=\frac{(n+1)!*(n+2)*(n+3)}{(n+1)!}=(n+2)*(n+3)=n^2+5n+6$